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Reliability analysis of a repairable system without being repaired “as good as new”
Microelectron. Reliab., Vol. 34, No. 2, pp. 357-360, 1994. Printed in Great Britain.
0026-2714/9456.00 + .00 © 1993 Pergamon Press Ltd
RELIABILITY ANALYSIS OF A REPAIRABLE SYSTEM W I T H O U T BEING REPAIRED "AS GOOD AS NEW" W U SHAO=MING, HUANG REN a n d WAN DE-JUN
Department of Instrument Science and Engineering, Southeast University, Nangjing 210 018, People's Republic of China
(Receivedfor publication 7 July 1992) Abstract--This paper considers a repairable system which is a two-unit priority standby redundant system. If units fail, they cannot be repaired to be "as good as new". Under this assumption, by using a geometric process and the method of a supplementary variable, some reliability indices are derived.
1. INTRODUCTION
ately and unit 2 is replaced in operation. Unit 2
The reliability of a standby redundant system, composed of two units in which one is given priority for operation and repair has been studied. The above system has been discussed [1] using Markov renewal processes, and assuming a failure system would be
operates until unit 1 is finished being repaired. On the other hand, if unit 2 is under repair and unit 1 fails in operation, then unit 1 is given priority for repair, whether the repair of unit 2 is finished or not. It is assumed that each switchover is perfect and each switchover time is instantaneous.
"as good as new". However, a repaired failure system is always different from a new one. For example, after repair, the operational time of a machine will become shorter and shorter with the increasing n u m b e r of repair times, so the total operational time or the total life of the machine must be finite. On the other hand, in view of the ageing and accumulative wear, the repair time will become longer and longer and tend towards infinity, i.e. finally, the machine is non-repairable [2]. On the basis of this, and by using geometric processes and a supplementary variable method, some explicit expressions are derived. Considering this case, that two units can not be repaired "as good as new", it is obvious that the results of Ref. [1] are particular examples of the method proposed,
2. Assuming that the repaired unit may not be "as good as new", let X~3 and Y~) be the life and the repair time of unit i during the k t h period, respectively (where period is the time that a unit spends from the beginning of operation or in standby to the end of finishing the repair), where i = 1,2 and k = 1, 2 . . . . . The distribution functions of X~,l) and YCk')are
F~l)(t)=Fl(kt)= 1 - e x p
respectively, where ai > 1, 0 < bl < 1. The distribution functions of X~,2) and y~2) are
Definition 1. Geometric processes [2] Given a sequence of random {X,, n = 1, 2 . . . . }, for some a > 0, then
)
a~-lgl(ak-lx)dx (f0 ( f0 ) G~)(t) = GI (kt) = 1 - e x p bk- ~/A(bk- ly) dy ,
F~2)(t) = F2(kt ) = 1 - exp(-a2*-12:) variables
{a n- IXn, n = 1, 2 . . . . }
G~,2)(t) = G2(kt) = 1 - e x p ( - b k- l/h), respectively, where a: > 1, 0 < b2 < 1.
forms a renewal process, and a is the parameter of the geometric process. Obviously, (1) if a > 1, then {X,, n = 1, 2 . . . . } is stochastically decreasing and converges to zero with a probability 1. (2) If 0 < a < 1, then {Xn, n = 1, 2 . . . . } is stochastically increasing and converges to infinity with a probability 1. ( 3 ) I f a = 1, then the geometric process {Xn, n = 1, 2 . . . . } is a renewal process, 2. MODEL 1. Let a system be composed of two units in which unit 1 is given priority to operate and unit 2 is not. If unit 1 does not work, it undergoes repair immedi-
3. SYSTEM ANALYSIS According to the above model, {X~°, i = 1, 2 . . . . } is stochastically decreasing and { Y~3, i = 1, 2 . . . . } is stochastically increasing. With the lapse of time, the life of the units will be shorter and shorter, and the repair time will be longer and longer after every repair.
Definition 2 Let N(t) be the system state at time t, where N(t) = 0: At time t, unit 1 is operating, unit 2 is in standby. N(t) = 1: At time t, unit 1 is being repaired, unit 2 is operating. N(t) = 2: At time t, unit 1 is 357
Wu SrtAO-MaNGet al.
358
operating, unit 2 is being repaired. N(t) = 3: At time t, unit 1 is being repaired, unit 2 is waiting for repair. Obviously, {N(t), t t>0} is a random process, whose state space is E = {0, 1, 2, 3}. From the above assumption, {N(t),t i>0} is not a Markov process, Therefore, we may introduce a supplementary variable: let /~(t) be the period of unit i at time t, respectively, X(t) be the life of unit l at time t during period 11(t), Y(t) be the repair time of unit 1 at time t during period Ii(t), then {N(t), ll(t), I2(t), X(t), Y(t)} is a higher dimensional Markov process. We define the following state probability: P0kj(t, x) = P { N ( t ) = 0, 11(t) = k, I2(t) =j, x <~X(t) ~
plkj(t,y)=e{N(t)=l,I~(t)=k,
12(t)=j,y<~Y(t)
Plkj(t, 0) =
a~ - 12~(a~ - ly) pokj(t, x) dx
p2u(t, 0) = 0 /'~ pz~j(t, 0) = J 0 bkl-21~l(b~-2y)P3k-~J(t'y)dy'
fo a,kP3kj(t, O) =
(8) (9)
k/>2
(10)
21(al ~y)p2kj(t,x)dx
(11)
l
k-
Ptkj(O, Y) = 0, where l = l, 3, ptkj(O, X) = 0, where i = 0, 2. Taking Laplace-Stieltjes transforms on both sides of equation (3), we have [d-~ + s + b k~- I #~t"bkl IYl"+a~-'221P*kj(S'y)=O'
~
(12)
P2kj(t,x)=P{N(t)=2, lj(t)=k, lE(t)=j,x<~X(t )
where
~
f(t)e-"dt.
f*(s) = Denote
~
~l)(y) = l - G~,l)(y), then we have
pokj(t + At, x + At)
p~'kj(S, y) = CIkj~I)(y)exp(--sY -- a~2- 122Y), (13)
=Pokj(t, X)[1 -- a k- 121(ak-IX) At]
where C~kj will be given at the end of this section. For the same reason, by using initial conditions and taking Laplace-Stieltjes transforms on both sides of equations (1), (2), (4) and (5) denote ~'~,l~(x)=l-F~,l~(x),thenwehave
+p~j_~(t, x)bJ2-E/.t2At +0(At), namely It3 p X +-~-X O +~ a l k-I 21(atk - l x ) ] 0kj(t,x)
=bJ2-2p2p2ky i(t,X), j~>2.
(1)
In the same way, we have ~ + ~ x x + a l k- l21(alk - lx)
Okl(t, x ) = 0
(2)
p*kj(s, x) = [Cokjexp(-sx) - C2kj- l X exp(--sx -- bJ2 2/ZEX)]~'~l)(x), j/> 2.
(14)
p *kl (S, X) = Co~l exp(-- SX )~kl)(X )
05)
p*j(s,x)=C2kjexp(-sx-b~
(16)
lp2x)~l)(x)
[dO ~ -~ +-~y +b~-~#~(b~-'y)+a~-12: ~k~(t,y)=O
p*k~(s, y ) = [C3k~e x p ( - sy ) - Clgj × exp(--sy --a~ ~2~y)]G),l)(y) (17)
(3) IO ~ ]p ~+~x+a~-12~(ak-~x)+b~-~ta~ 2~)(t,x) = 0
Taking Laplace-Stieltjes transforms on both sides of equations (6-11) and by using equations (13-17), denote
gk(s) =
(4)
+-~y + b~-'l~l(b~-~x) p3~(t,y) = aJ2- ~2~pm~(t, y).
(5)
The boundary and initial conditions are
fO
Pokj(to)=
e -sy dG~,~(Y),
f~
fk(s) =
po.(t,O)=f(t),
I: Jo
e st dF~)(y),
we have
po.~(t,0 ) = 0
(6)
b~-21zm(b~ :y)ptk_~(t,y)dy, k t> 2 (7)
Co~l=l,
Co.j=0,
j~>2
Cokl-~-Clk_llgk_l(S"[-,~,2),
(18)
k >~2 (19) Cok~=C2k~_l +C~k_~ygk_~(s +a~-t2:)k, j > . 2 (20)
A repairable system
359
(21)
Clk, = Coklfk(s)
= ~ r.l -gk(s + aJ2-122)~j_12 r
cikj= Coke(s) + c2~.lA(s +b~-2~2), j >_.2
~-l L
s
-b a~-lj.2
u2
2".-lkj
"-I
(22) C21j= 0
(23)
Cz~j= C3k-ljgk-l(S)+ Clk-ljgk- 1 x(s+a~-122),
ki>2
C3ky= C2kj+ CzkJ~(S + b~- llh).
(24) (25)
4. RELIABILITY
+
A(t)= j,k=l
[f;
Pokj(t,x) dx +
fo
Let M represent the expected number of visits to the system failure per unit in the steady state, then n = lim MI(t) = iim (sW~ (s)). (30) t~o t t~0 3. System reliabifity. In order to obtain system reliability, state 3 is taken for an absorbing state, then all P3kj(t, x) in Section 2 can be deleted. Let R(t) be system reliability, then, we have
p*kj(s,x)dx +
j,k=l
+
l
+
Taking the Laplace-Stieltjes transforms of both sides of equation (26) and using equations (13-16), we have A*(s)=
k=~[fo °° Pokj(t,x)dx
R(t)=j~
pLkj(t,y)dy
+J0 p~j(t,x)dxJ. (26) p~,j(s,y)dy
(29) J
1. Availability. DenoteA(t) asthesysteminstantaneous availability and, according to Ref. [3], we have
+
fo
plk](t,y) dy +
f;
PEki(t,x)dx
p*kj(S, X) = Cokjexp(--sx)F~l)(x) p*kj(s,y)=Clkjexp(--sy--a~-122y)~l'(y) p*j (s, x) = 0,
]
(31) (32) (33) (34)
where
Cokl=[If_,(s)g,_l(S +22), k >.2
C011=l,
i=2
f;
p~j(s, x) dx
]
C0~j=0, j~>2,
= ~ [1--fk(s+b~-lp2 ) j,k=l
S-+-~2:~
CHI=fI(s)
(35) (36)
k Cz~j
Cl,i=A(s) I-If_,(s)g,_l(S+22),
k>~2 (37)
i=]
+ 1--gk(s+a~-lA2)s +--a~--- 12--2- Clkjj]
Clkj = 0, j~>2.
Taking the Laplace-Stieltjes transforms on both sides ofequation (31) and using equations (32-38), we have
+ ~ ~ [1--fk(S) ~ j=2k=l L s (~Okj
-~- l - f k ( s ) S
Cok I .
(38)
+joP'kj(S'x)dxJ
(27)
2. Failure frequency of the system. Let Wf(t) represent the failure frequency of the system, My(t) represent the mean failure numbers of the system during (0, t]. According to Ref. [3], we have Wf(t)=~[f;Plkj(t,y)a~-122dY i, k= l
=
.1 -- (s) k=2 +
~
f(s)gi(s
"~-'~21
i=l
1--gk(s Jr /~2)fk(s) k-I q 1-I f(s)gi(s + 22) S J r '~2
i=l
1-fl(s) 1 - g , ( s +22) -t - + s s +22
/
(39)
+Ji~pz~j(t,x)a~-lal(a~-lx)dx].(28/ Taking the Laplace-Stieltjes transforms on both sides of equation (28), we have
W~(s)=
p*kj(s,y)a{-]22dy j,k=l
+~;*pz~j(S,X)alk-~ ] 2 k-I ' l (x )ad x
5. DISCUSSION This paper is based on the assumption that a~ and b~ do not equal 1, where i = 1,2. In the view of the geometric process, if a~ = b~= 1, the {X~~, n = 1, 2 . . . . }, is an i.i.d, sequence, as is the {Y°~,n . .1,2, . . }, and the period of units will lose its meanings, the failed units will be repaired "as good
WU SI-IAO-MINGet al.
360
as new", then all k, j, at and bi in Section 2 can be deleted. By this, the initial conditions must be considered in equation (7), i.e.
po(t,O)=f)#lO,)p,(y)dy+t$(t),
By using equations (29), (30), (45) and (46), we have M = lim sW~(s) s~0
(40)
=limsI1-_gs(S~_22)22 1 1 s~o + 2 -f(s)g(s) f(s ) - f(s )g(s ) f(s + #2)
equation (20) will become:
Co = C2 + C,g(s + 22) + 1,
4
(41)
[1 -f(s)g(s)][1 - f ( s + #2)g(s + 22)]
x f(s + #2)]
where
d 21#, [1 - g(22)]
g(s) = L - e-SY dG]°(Y)' do
denote
(21 + #j ) [1 - f ~ 2 ) g ( 2 2 ) ]
(47)
where
f(s) =
e-SX dF~l)(x)'
1 -- =
using equations (22), (24) and (25), we have
2j
Cl = Cof(s) - C2f(s + #2)
(42)
C2 = C 2 g ( s ) - C~g(s + 22) C3 = CI + C2f(s + #2).
(43) (44)
Solving equations (41-44), we have equations 1 C0--1-f(s)g(s)'
f(s)g(s) --f(s)g(s)(s + 22) Cl = [1 --f(s)g(s)][l --f(s + #2)g(s + )~2)] '
(45)
f(s) --f(s)g(s)f(s + #2) C2 = [1 -f(s)g(s)][l - f ( s + #2)g(s + 22)] '
f(s) C3 = 1 - f ( s ) g ( s ) "
(46)
1 #J
fo
x dF~l~(x),
Y dG]l)(y).
Note that the result of equation (47) is equal to equation (22) in Ref. [1]. In the same way, if we let ai = b~ = 1 (i = 1, 2), we will have the same results as Ref. [1]. To sum up, this paper extends Osaki's model. Users will find that the model in this paper corresponds more closely with actual situations. REFERENCES
1. T. Nagakawa and S. Osaki, Microelectron. Reliab. 15, 309 (1975). 2. Lam Yeh, Adv. Appl. Prob. 20, 479-482 (1988). 3. Shi Ding-hua, ACTA Math. Appl. Sinica 1, 101-110 (1985).
0026-2714/9456.00 + .00 © 1993 Pergamon Press Ltd
RELIABILITY ANALYSIS OF A REPAIRABLE SYSTEM W I T H O U T BEING REPAIRED "AS GOOD AS NEW" W U SHAO=MING, HUANG REN a n d WAN DE-JUN
Department of Instrument Science and Engineering, Southeast University, Nangjing 210 018, People's Republic of China
(Receivedfor publication 7 July 1992) Abstract--This paper considers a repairable system which is a two-unit priority standby redundant system. If units fail, they cannot be repaired to be "as good as new". Under this assumption, by using a geometric process and the method of a supplementary variable, some reliability indices are derived.
1. INTRODUCTION
ately and unit 2 is replaced in operation. Unit 2
The reliability of a standby redundant system, composed of two units in which one is given priority for operation and repair has been studied. The above system has been discussed [1] using Markov renewal processes, and assuming a failure system would be
operates until unit 1 is finished being repaired. On the other hand, if unit 2 is under repair and unit 1 fails in operation, then unit 1 is given priority for repair, whether the repair of unit 2 is finished or not. It is assumed that each switchover is perfect and each switchover time is instantaneous.
"as good as new". However, a repaired failure system is always different from a new one. For example, after repair, the operational time of a machine will become shorter and shorter with the increasing n u m b e r of repair times, so the total operational time or the total life of the machine must be finite. On the other hand, in view of the ageing and accumulative wear, the repair time will become longer and longer and tend towards infinity, i.e. finally, the machine is non-repairable [2]. On the basis of this, and by using geometric processes and a supplementary variable method, some explicit expressions are derived. Considering this case, that two units can not be repaired "as good as new", it is obvious that the results of Ref. [1] are particular examples of the method proposed,
2. Assuming that the repaired unit may not be "as good as new", let X~3 and Y~) be the life and the repair time of unit i during the k t h period, respectively (where period is the time that a unit spends from the beginning of operation or in standby to the end of finishing the repair), where i = 1,2 and k = 1, 2 . . . . . The distribution functions of X~,l) and YCk')are
F~l)(t)=Fl(kt)= 1 - e x p
respectively, where ai > 1, 0 < bl < 1. The distribution functions of X~,2) and y~2) are
Definition 1. Geometric processes [2] Given a sequence of random {X,, n = 1, 2 . . . . }, for some a > 0, then
)
a~-lgl(ak-lx)dx (f0 ( f0 ) G~)(t) = GI (kt) = 1 - e x p bk- ~/A(bk- ly) dy ,
F~2)(t) = F2(kt ) = 1 - exp(-a2*-12:) variables
{a n- IXn, n = 1, 2 . . . . }
G~,2)(t) = G2(kt) = 1 - e x p ( - b k- l/h), respectively, where a: > 1, 0 < b2 < 1.
forms a renewal process, and a is the parameter of the geometric process. Obviously, (1) if a > 1, then {X,, n = 1, 2 . . . . } is stochastically decreasing and converges to zero with a probability 1. (2) If 0 < a < 1, then {Xn, n = 1, 2 . . . . } is stochastically increasing and converges to infinity with a probability 1. ( 3 ) I f a = 1, then the geometric process {Xn, n = 1, 2 . . . . } is a renewal process, 2. MODEL 1. Let a system be composed of two units in which unit 1 is given priority to operate and unit 2 is not. If unit 1 does not work, it undergoes repair immedi-
3. SYSTEM ANALYSIS According to the above model, {X~°, i = 1, 2 . . . . } is stochastically decreasing and { Y~3, i = 1, 2 . . . . } is stochastically increasing. With the lapse of time, the life of the units will be shorter and shorter, and the repair time will be longer and longer after every repair.
Definition 2 Let N(t) be the system state at time t, where N(t) = 0: At time t, unit 1 is operating, unit 2 is in standby. N(t) = 1: At time t, unit 1 is being repaired, unit 2 is operating. N(t) = 2: At time t, unit 1 is 357
Wu SrtAO-MaNGet al.
358
operating, unit 2 is being repaired. N(t) = 3: At time t, unit 1 is being repaired, unit 2 is waiting for repair. Obviously, {N(t), t t>0} is a random process, whose state space is E = {0, 1, 2, 3}. From the above assumption, {N(t),t i>0} is not a Markov process, Therefore, we may introduce a supplementary variable: let /~(t) be the period of unit i at time t, respectively, X(t) be the life of unit l at time t during period 11(t), Y(t) be the repair time of unit 1 at time t during period Ii(t), then {N(t), ll(t), I2(t), X(t), Y(t)} is a higher dimensional Markov process. We define the following state probability: P0kj(t, x) = P { N ( t ) = 0, 11(t) = k, I2(t) =j, x <~X(t) ~
plkj(t,y)=e{N(t)=l,I~(t)=k,
12(t)=j,y<~Y(t)
Plkj(t, 0) =
a~ - 12~(a~ - ly) pokj(t, x) dx
p2u(t, 0) = 0 /'~ pz~j(t, 0) = J 0 bkl-21~l(b~-2y)P3k-~J(t'y)dy'
fo a,kP3kj(t, O) =
(8) (9)
k/>2
(10)
21(al ~y)p2kj(t,x)dx
(11)
l
k-
Ptkj(O, Y) = 0, where l = l, 3, ptkj(O, X) = 0, where i = 0, 2. Taking Laplace-Stieltjes transforms on both sides of equation (3), we have [d-~ + s + b k~- I #~t"bkl IYl"+a~-'221P*kj(S'y)=O'
~
(12)
P2kj(t,x)=P{N(t)=2, lj(t)=k, lE(t)=j,x<~X(t )
where
~
f(t)e-"dt.
f*(s) = Denote
~
~l)(y) = l - G~,l)(y), then we have
pokj(t + At, x + At)
p~'kj(S, y) = CIkj~I)(y)exp(--sY -- a~2- 122Y), (13)
=Pokj(t, X)[1 -- a k- 121(ak-IX) At]
where C~kj will be given at the end of this section. For the same reason, by using initial conditions and taking Laplace-Stieltjes transforms on both sides of equations (1), (2), (4) and (5) denote ~'~,l~(x)=l-F~,l~(x),thenwehave
+p~j_~(t, x)bJ2-E/.t2At +0(At), namely It3 p X +-~-X O +~ a l k-I 21(atk - l x ) ] 0kj(t,x)
=bJ2-2p2p2ky i(t,X), j~>2.
(1)
In the same way, we have ~ + ~ x x + a l k- l21(alk - lx)
Okl(t, x ) = 0
(2)
p*kj(s, x) = [Cokjexp(-sx) - C2kj- l X exp(--sx -- bJ2 2/ZEX)]~'~l)(x), j/> 2.
(14)
p *kl (S, X) = Co~l exp(-- SX )~kl)(X )
05)
p*j(s,x)=C2kjexp(-sx-b~
(16)
lp2x)~l)(x)
[dO ~ -~ +-~y +b~-~#~(b~-'y)+a~-12: ~k~(t,y)=O
p*k~(s, y ) = [C3k~e x p ( - sy ) - Clgj × exp(--sy --a~ ~2~y)]G),l)(y) (17)
(3) IO ~ ]p ~+~x+a~-12~(ak-~x)+b~-~ta~ 2~)(t,x) = 0
Taking Laplace-Stieltjes transforms on both sides of equations (6-11) and by using equations (13-17), denote
gk(s) =
(4)
+-~y + b~-'l~l(b~-~x) p3~(t,y) = aJ2- ~2~pm~(t, y).
(5)
The boundary and initial conditions are
fO
Pokj(to)=
e -sy dG~,~(Y),
f~
fk(s) =
po.(t,O)=f(t),
I: Jo
e st dF~)(y),
we have
po.~(t,0 ) = 0
(6)
b~-21zm(b~ :y)ptk_~(t,y)dy, k t> 2 (7)
Co~l=l,
Co.j=0,
j~>2
Cokl-~-Clk_llgk_l(S"[-,~,2),
(18)
k >~2 (19) Cok~=C2k~_l +C~k_~ygk_~(s +a~-t2:)k, j > . 2 (20)
A repairable system
359
(21)
Clk, = Coklfk(s)
= ~ r.l -gk(s + aJ2-122)~j_12 r
cikj= Coke(s) + c2~.lA(s +b~-2~2), j >_.2
~-l L
s
-b a~-lj.2
u2
2".-lkj
"-I
(22) C21j= 0
(23)
Cz~j= C3k-ljgk-l(S)+ Clk-ljgk- 1 x(s+a~-122),
ki>2
C3ky= C2kj+ CzkJ~(S + b~- llh).
(24) (25)
4. RELIABILITY
+
A(t)= j,k=l
[f;
Pokj(t,x) dx +
fo
Let M represent the expected number of visits to the system failure per unit in the steady state, then n = lim MI(t) = iim (sW~ (s)). (30) t~o t t~0 3. System reliabifity. In order to obtain system reliability, state 3 is taken for an absorbing state, then all P3kj(t, x) in Section 2 can be deleted. Let R(t) be system reliability, then, we have
p*kj(s,x)dx +
j,k=l
+
l
+
Taking the Laplace-Stieltjes transforms of both sides of equation (26) and using equations (13-16), we have A*(s)=
k=~[fo °° Pokj(t,x)dx
R(t)=j~
pLkj(t,y)dy
+J0 p~j(t,x)dxJ. (26) p~,j(s,y)dy
(29) J
1. Availability. DenoteA(t) asthesysteminstantaneous availability and, according to Ref. [3], we have
+
fo
plk](t,y) dy +
f;
PEki(t,x)dx
p*kj(S, X) = Cokjexp(--sx)F~l)(x) p*kj(s,y)=Clkjexp(--sy--a~-122y)~l'(y) p*j (s, x) = 0,
]
(31) (32) (33) (34)
where
Cokl=[If_,(s)g,_l(S +22), k >.2
C011=l,
i=2
f;
p~j(s, x) dx
]
C0~j=0, j~>2,
= ~ [1--fk(s+b~-lp2 ) j,k=l
S-+-~2:~
CHI=fI(s)
(35) (36)
k Cz~j
Cl,i=A(s) I-If_,(s)g,_l(S+22),
k>~2 (37)
i=]
+ 1--gk(s+a~-lA2)s +--a~--- 12--2- Clkjj]
Clkj = 0, j~>2.
Taking the Laplace-Stieltjes transforms on both sides ofequation (31) and using equations (32-38), we have
+ ~ ~ [1--fk(S) ~ j=2k=l L s (~Okj
-~- l - f k ( s ) S
Cok I .
(38)
+joP'kj(S'x)dxJ
(27)
2. Failure frequency of the system. Let Wf(t) represent the failure frequency of the system, My(t) represent the mean failure numbers of the system during (0, t]. According to Ref. [3], we have Wf(t)=~[f;Plkj(t,y)a~-122dY i, k= l
=
.1 -- (s) k=2 +
~
f(s)gi(s
"~-'~21
i=l
1--gk(s Jr /~2)fk(s) k-I q 1-I f(s)gi(s + 22) S J r '~2
i=l
1-fl(s) 1 - g , ( s +22) -t - + s s +22
/
(39)
+Ji~pz~j(t,x)a~-lal(a~-lx)dx].(28/ Taking the Laplace-Stieltjes transforms on both sides of equation (28), we have
W~(s)=
p*kj(s,y)a{-]22dy j,k=l
+~;*pz~j(S,X)alk-~ ] 2 k-I ' l (x )ad x
5. DISCUSSION This paper is based on the assumption that a~ and b~ do not equal 1, where i = 1,2. In the view of the geometric process, if a~ = b~= 1, the {X~~, n = 1, 2 . . . . }, is an i.i.d, sequence, as is the {Y°~,n . .1,2, . . }, and the period of units will lose its meanings, the failed units will be repaired "as good
WU SI-IAO-MINGet al.
360
as new", then all k, j, at and bi in Section 2 can be deleted. By this, the initial conditions must be considered in equation (7), i.e.
po(t,O)=f)#lO,)p,(y)dy+t$(t),
By using equations (29), (30), (45) and (46), we have M = lim sW~(s) s~0
(40)
=limsI1-_gs(S~_22)22 1 1 s~o + 2 -f(s)g(s) f(s ) - f(s )g(s ) f(s + #2)
equation (20) will become:
Co = C2 + C,g(s + 22) + 1,
4
(41)
[1 -f(s)g(s)][1 - f ( s + #2)g(s + 22)]
x f(s + #2)]
where
d 21#, [1 - g(22)]
g(s) = L - e-SY dG]°(Y)' do
denote
(21 + #j ) [1 - f ~ 2 ) g ( 2 2 ) ]
(47)
where
f(s) =
e-SX dF~l)(x)'
1 -- =
using equations (22), (24) and (25), we have
2j
Cl = Cof(s) - C2f(s + #2)
(42)
C2 = C 2 g ( s ) - C~g(s + 22) C3 = CI + C2f(s + #2).
(43) (44)
Solving equations (41-44), we have equations 1 C0--1-f(s)g(s)'
f(s)g(s) --f(s)g(s)(s + 22) Cl = [1 --f(s)g(s)][l --f(s + #2)g(s + )~2)] '
(45)
f(s) --f(s)g(s)f(s + #2) C2 = [1 -f(s)g(s)][l - f ( s + #2)g(s + 22)] '
f(s) C3 = 1 - f ( s ) g ( s ) "
(46)
1 #J
fo
x dF~l~(x),
Y dG]l)(y).
Note that the result of equation (47) is equal to equation (22) in Ref. [1]. In the same way, if we let ai = b~ = 1 (i = 1, 2), we will have the same results as Ref. [1]. To sum up, this paper extends Osaki's model. Users will find that the model in this paper corresponds more closely with actual situations. REFERENCES
1. T. Nagakawa and S. Osaki, Microelectron. Reliab. 15, 309 (1975). 2. Lam Yeh, Adv. Appl. Prob. 20, 479-482 (1988). 3. Shi Ding-hua, ACTA Math. Appl. Sinica 1, 101-110 (1985).